Is the Markov property assumed in the forward algorithm?
In general, to formally state that the Markov property holds, you need to have $P(
x_t \mid x_{t-1:1}) = P(x_t \mid x_{t-1})$.

So, you cannot conclude only from $P(q_{t-1} = i, q_t=j, o_1^{t-1}, o^t|\lambda)$ that the Markov property holds, because  $P(q_{t-1} = i, q_t=j, o_1^{t-1}, o^t|\lambda)$ is the joint probability of $q_{t-1} = i, q_t=j, o_1^{t-1}$ and $o^t$ given $\lambda$.

Nonetheless, the theory of hidden Markov model often assumes that a few properties (including the Markov property) hold


the Markov property: $$P(q_{t+1} \mid q_{t}) = P(q_{t+1} \mid q_{t:1}),$$ where $q_t$ is the hidden state at time step $t$ and $q_{t:1} = q_t, q_{t-1}, \dots, q_1$.
the stationarity property: $$P(q_{t_1 + 1} \mid q_{t_1}) = P(q_{t_2 + 1} \mid q_{t_2}),$$ for any time step $t_1$ and $t_2$. In other words, the state transition probabilities are independent of the actual time at which the transitions takes place.
the output independence property: $$P(o^{T:1} \mid q_{T:1}) = \prod_{t=1}^T P(o^t \mid q_t , \lambda)$$
where $o^{T:1} = o^T, o^{T-1}, \dots, o^1$. In other words, this is the assumption that the output at time step $t$, $o^t$, is independent of the outputs at previous time steps.